Basically every assembler, furnace and even the mining drills use productivity modules 3. My borders are fortified with 3 rows of walls, 1 row of laser towers and fully covered with roboports and repair bots (see thick blue lines in screenshot 1). At the moment thats 48 roboports and 1500 Laser turrets. Therefore I need A LOT of energy, but I dont like to waste masses of coal, so I power everything completely with solar panels and accumulators. At the moment I have 2400 Solar modules and 2000 accumulators, in the screenshot you can see a part of the complex. I have spended quite a bit of time on the layout.

I can't wait for the blueprints implementation, I am sick of building this things manually ^^

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Last edited by BurnHard on Wed Dec 25, 2013 7:35 pm, edited 1 time in total.

You can be more space efficient than that. Substations have a line reach just long enough that it's possible to make it possible to cover a field without overlap or blind spots. Because of this you can stack 18 accu's, 56 solar panels and 3 substations in a single, self contained block, which can be repeated as long as there is construction space left and it doesn't matter if it's oriented north-south or east-west. It does lack sufficient storage capacity to power the factory at night, nevermind the defenses, but you can just drop in some accumulator fields of the same size in to cover that problem.

Park a large electric tower close by and you can hook the entire array up to your electric grid. The only bad thing is that you have to go around the entire array as you can't cut straight through it.

Hazard wrote:Substations have a line reach just long enough that it's possible to make it possible to cover a field without overlap or blind spots.

If you look closely, I have 2x2 substations combined. I may try a 3x3 or 4x4 variation next time

Hazard wrote:The only bad thing is that you have to go around the entire array as you can't cut straight through it.

Thats the exact reason why I build it that way. I like to have enough space to walk through, hence I placed accumulators in the edges of the singe large elements near the large power poles. For me the ratio of accus and solar modules work very good as well.

I always went the opposite, using just lv2 power poles and solar panels. I think this way gives you better density, but might be more resource intensive.

That is me, standing in a square 8 poles x 8 poles wide with 8 solar panels spiraling out from the center of each pole (512 panels in a 8x8x8 poles grid). the 2 squares on my map the 2 different directions you could lay out he grid.

The more massive you build these solar panel grid structures, the smaller the surface area in comparison to the total number of solar panels used. (You have to round off the edges and make a sphere to achieve the optimum perimeter). This is (theoretically) the most efficient layout for defending a large solar farm with laser turrets. (least amount of laser turrets vs. most panels protected on the inside of the bubble)

would you be interested in renting me out some power so i don't have to build something this massive LOL, very nice job, i love coming to the forums and seeing someone do something massive like this with the game. kudos again man.

Pandamonium wrote:
Q1:
If I were to change over from steam --> solar/accus, what's the best ratio of solar : accu's to use?
I've forgotten enough of my physics to forget how to calculate it
Q2:
currently have 8 steam/8 burners "power plants".
how many solar panels/accu combos would be require to place one of my "power plants"?

Q1: Well, it depends. 1:1 is ok, but ususally I build more accus. Its really more trial and error at the moment, see anwer 2. Until I have really a LOT of solar and accus I alyways leave some of the steam engines on "standby" by disconnecting eg the water pipe.

Q2: Minimum 10 times the steam engines, realistic 15-20 times. When I play a game, I build a bunch of solars and accus and just look after the first night, when were my accus empty and how fast where they recharged in the following day. leave some of the steam engines connected (you'll need power in the night somehow). If the accus are too fast empty, I build more accus. If the aren't loaded full the following day I build more solar panels. At the moment its 1:1 more ore less.

In the endgame I dont want to waste that much space on solar panels any more and the ratio gets more 1:2... So I have enough capacity for some nights, even if the solar panels dont charge full one day.

At the complete beginning you can build just only solar fields, so than at the day your complete energy comes from solar, and at night the steam engines would work. Accus are not really needed in the beginning

This uses three individual blueprints:
1) The Solar arrays... These are 6x6 solar farms with the inner 4 missing, replaced with 1 big pole, 1 substation (PERFECTLY covering the solar array), and 7 accus to take up the rest of the space.
2) The full-block Accumulators around the edges. 7x7 Accumulators with a substation in the middle. Boring, but effective.
3) The Middle/Corner blocks: 7x7 accumulators, with a big power pole and a roboport. These *perfectly* link the entire array into a single robo zone, for quick self-building via a blueprint.

The whole array is a 5x5 block of #1 solar arrays with 5x #2 and 2x #3 on each side. Sadly the sides don't match up perfectly, but what can you do ^_^;

I did actually create a blueprint of the entire gigantic array, I'm just can't find a place to put a second one... It's perfectly square, auto-buildable, and tileable, at 1.4k Accumulators, 800 Solar Panels, 53 substations, 34 Big Poles, and 8 Roboports.

Pandamonium wrote:
Q1:
If I were to change over from steam --> solar/accus, what's the best ratio of solar : accu's to use?
I've forgotten enough of my physics to forget how to calculate it

Warning: some math here, if you're interested in answer only skip to last paragraph.

So I've taken the time to check how much electricity is used in one night (that is in time when accumulators start discharging and start charging again). It seems that rough amount of energy needed to last the night is:

per day. Each accumulator can contain 5 MJ of energy. So this means that it takes 5 solar panels to maintain 3 accumulators (given that you never run out of energy during the day). Also since you can only use half of what solar panel produces to power your devices, for each steam generator you need

Answer:
You need 17 solar panels for each steam engine running at max capacity.
Or more precisely you need power_consumption_in_w / 30 kW solar panels.
For each 5 solar panels you need 3 accumulators.
Keep in mind that these numbers are approximate, you may want to have a buffer (for laser turrets if nothing else).

So I did some math to figure out how many accumulators and solar panels you need per KW.

I set up a radar station that uses 310KW (says it uses 300 but power poles say differently. I needed 8 solar panels and 6 accumulators to power it at max power and I had barely any power storage left before the solar panels were fully operational.

My results are that you need .0258 solar panels and .0193 accumulators for every kilowatt you consume.

The idea is that you want to have the amount of capacitors sufficient enough to never discharge below 50%. If they do, just add more {blocks} <- edit.
The ⅔ ratio allows panels to restore ~85% accumulator capacity during the daytime while still powering anything that discharged said capacitors.

This is a bigger block

Last edited by User_Name on Tue Nov 11, 2014 2:02 pm, edited 1 time in total.

Well don't want to sound rude but only because you do it like that doesn't mean it's the optimal way does it? I've posted all my math on how I calculated all that and unless you can prove me wrong with numbers I'll stick to saying that my way is a lot more optimal (in terms of resources and space needed to power up a base) than yours. It is of course not perfect since it was only an approximation that you need consumption_in_W x 100 in accumulator capacity but it's accurate enough for most cases.

I calculated the ratio in OPs image and it seems to me that it's pretty close to optimal layout while keeping some buffer in case of unexpectedly many enemies during the night.

sinsiliux wrote:Well don't want to sound rude but only because you do it like that doesn't mean it's the optimal way does it? I've posted all my math on how I calculated all that and unless you can prove me wrong with numbers I'll stick to saying that my way is a lot more optimal (in terms of resources and space needed to power up a base) than yours. It is of course not perfect since it was only an approximation that you need consumption_in_W x 100 in accumulator capacity but it's accurate enough for most cases.

I calculated the ratio in OPs image and it seems to me that it's pretty close to optimal layout while keeping some buffer in case of unexpectedly many enemies during the night.

1) The night has nothing to do with solar panels to capacitors ratio, or solar panels at all for that matter. Its just the capacitor count what matters. You either have enough, or you don't.
2) Having more panels than you need cannot be "more optimal in terms of resources and space needed".

The goal is to find lowest possible ratio that still works reliably.
⅔ works, if you have enough capacitors to not drop below 50% charge at night (it restores about 85% of maximum accumulator capacity).

If you make some work and search back in this board you'll find some more examples about optimal placing of solar/accu, calculations about the optimal ratio between that, calculations about how much power you need for the night etc.

User_Name wrote:
1) The night has nothing to do with solar panels to capacitors ratio, or solar panels at all for that matter. Its just the capacitor count what matters. You either have enough, or you don't.

It does, you need enough accumulators to last through the night and enough solar panels to charge those accumulators during the day.

User_Name wrote:
2) Having more panels than you need cannot be "more optimal in terms of resources and space needed".

The goal is to find lowest possible ratio that still works reliably.
⅔ works, if you have enough capacitors to not drop below 50% charge at night (it restores about 85% of maximum accumulator capacity).

That's where you are wrong. It's not a matter of minimizing ratio of solar panels vs accumulators. It is a matter or minimizing total count of accumulators and total count of solar panels. For example let's say you need 1 GJ of electricity to last through the night, then:
1. Using your method you would have to build 1GJ / (85% - 50%) / 5MJ = 1 GJ / 0.35 / 5MJ = 572 accumulators. Using your ratio you would then need 572 * 2 / 3 = 382 solar panels.
2. Using my method you would need 1GJ / 5MJ = 200 accumulators. Using my ratio you would then need 200 * 5 / 3 = 334 solar panels.

User_Name wrote:
If you try to "minimize total count of accumulators" below the necessary amount you'll have blackout at night, no matter how many solar panels you have, one or thousand.

http://www.merriam-webster.com/dictionary/minimize. That's why you don't minimize below the necessary amount. That's what the whole math part was for - to calculate the minimal amount of accumulators needed to last through the night given your electricity usage. I don't know how many posts I can make trying to take different approach explaining the same thing...

User_Name wrote:
Bingo!
So, if ⅔ ratio is enough to charge "enough accumulators to last through the night", whats the point of having ratio 1.3? Half of your solar panels are unnecessary. How's that "optimizing"?

Because as I showed in my example you would need less of both - accumulators and solar panels. Did you even read my previous post?

Basic math and logic lesson:
Building more costs more
572 is more than 200
382 is more than 334
Both methods give same result - you always have enough energy
Using my method you can build less panels and accumulators, so my method is cheaper
If both methods provide same result but one is cheaper then cheaper method is better